Let ` y = x^(x^(x...oo))," then " (dy)/(dx) ` is equal to

Options

A. `yx^(y-1)`
B. `(y^(2))/(x(1-y log x))`
C. `(y)/(x (1 + y log x ))`
D. `None of these

Explanation

To solve the problem \( y = x^{x^{x^{...}}} \) (where the exponent goes to infinity), we need to find the derivative \( \frac{dy}{dx} \). Here’s a step-by-step solution:

### Step 1: Express \( y \) in terms of itself
Since \( y = x^{y} \) (because the exponent is \( y \) itself), we can write:
\[ y = x^y \]

### Step 2: Take the natural logarithm of both sides
Taking the logarithm of both sides gives us:
\[ \log y = \log(x^y) = y \log x \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides. Using implicit differentiation:
\[ \frac{d}{dx}(\log y) = \frac{d}{dx}(y \log x) \]
Using the chain rule on the left side and the product rule on the right side:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{dy}{dx} \log x + y \cdot \frac{1}{x} \]

### Step 4: Rearrange the equation
Rearranging the equation gives:
\[ \frac{1}{y} \frac{dy}{dx} - \frac{dy}{dx} \log x = \frac{y}{x} \]
Factoring out \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} \left( \frac{1}{y} - \log x \right) = \frac{y}{x} \]

### Step 5: Solve for \( \frac{dy}{dx} \)
Now, we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{y/x}{(1/y) - \log x} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{y^2}{x(1 - y \log x)} \]

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{y^2}{x(1 - y \log x)} \]